Državna Matura Math A common mistakes
Državna Matura Math A · Understand the traps and how to avoid them.
When solving an equation such as (x−1)/(x−3) = 2, the candidate multiplies both sides by (x−3) and reports x = 5 as the complete answer without ever stating x ≠ 3, so a solution that makes the original denominator zero is accepted or the excluded value is never mentioned.
School practice treats 'solve' as 'isolate x', and the candidate has automated the multiplication step without a habit of recording domain restrictions first. Because the excluded value rarely appears as the algebraic answer, nothing in the mechanics of the calculation flags it, so the check is simply never performed.
What to do instead
Before touching the equation, write the condition line 'denominator ≠ 0' and list every excluded value on the answer sheet. After finding roots, cross each one against that list and explicitly write 'x = 5, x ≠ 3' in the final answer. Practise on rational equations where the excluded value actually is a root of the multiplied equation, so the deletion becomes visible and memorable.
The candidate expands (a + b)² as a² + b², or writes √(a + b) = √a + √b, or cancels a term across a sum, e.g. reducing (x² + 3)/x to x + 3.
These come from over-generalising the distributive law: distribution works over multiplication, and the candidate applies the same visual pattern to powers, roots and addition. The notation looks linear, so the eye accepts the manipulation even though the operation is not linear.
What to do instead
Adopt a numerical sanity check: substitute a = 2, b = 3 into the original and the rewritten expression; if the values differ, the step is illegal. Drill the correct identities (a + b)² = a² + 2ab + b², a² − b² = (a − b)(a + b)) alongside deliberately wrong versions and have the candidate classify each as valid or invalid with a one-line justification.
When solving an inequality such as 3 − 2x > 7, the candidate divides by −2 and keeps the direction, giving x > −2 instead of x < −2; the same error appears when the inequality is multiplied by a negative expression.
The procedure for equations has been practised far more, and 'divide both sides by the coefficient' is executed as a reflex. Reversing the sign is a rule attached to one specific operation, and under time pressure the reflex fires before the rule is recalled.
What to do instead
Make sign reversal a written step, not a mental one: draw an arrow under the division line and write 'flip' at the moment of dividing by a negative number. Then always test the answer with a sample value (x = 0 should fail, x = −5 should satisfy 3 − 2x > 7) and check the test value against the original inequality, not the simplified one.
Given a quadratic function, the candidate reads the vertex directly from the coefficients, e.g. stating the vertex of y = 2x² − 8x + 5 is (−8, 5), or claims the parabola opens downward because the constant term is negative.
The candidate memorises 'the numbers in the formula are the features' from linear functions (y = kx + n), where the slope and intercept are literally the coefficients. Quadratics break that pattern, but the linear habit is older and stronger.
What to do instead
Force the vertex through completing the square or x = −b/(2a) every single time, and write the intermediate line. Then verify with two symmetric points: compute f(x_v + 1) and f(x_v − 1); if they are not equal, the vertex is wrong. Also state the sign of the leading coefficient separately as the opening direction before computing anything else.
For a logarithm, the candidate writes log(a + b) = log a + log b, or solves log(x − 1) + log(x + 3) = 1 by combining arguments and then accepts a root that makes one argument negative.
Logarithm laws are remembered as a list of transformations rather than as statements about products and quotients, so the product rule is applied to a sum. The domain restriction x > 0 is treated as a footnote rather than as part of the definition.
What to do instead
Write the definition log_a b = c ⇔ a^c = b at the top of every log problem and state the condition 'argument > 0' before solving. After obtaining roots, substitute each into the original arguments and discard any that give a non-positive value, showing the rejection explicitly. Contrast log(2·8) with log(2 + 8) numerically once, so the asymmetry is felt rather than memorised.
When using the sine rule or cosine rule in a triangle, the candidate pairs a side with the wrong opposite angle, or applies the cosine rule to find an angle and forgets that the result from arccos is the unique angle while the sine rule's arcsin can give an ambiguous obtuse alternative.
Diagrams are drawn roughly and the labelling is not fixed before substitution, so 'side a opposite angle A' becomes a guess. The ambiguous case is a genuine mathematical subtlety that is easily skipped when the candidate just wants a number.
What to do instead
Redraw the triangle and label vertices A, B, C and opposite sides a, b, c before any formula is chosen. Write the ratio a/sin A = b/sin B with the known pair in the first fraction. Whenever the sine rule is used to find an angle, explicitly check whether 180° − the found angle is also geometrically possible, and state which case applies.
In a geometric sequence the candidate confuses the common ratio r with the first term, or computes the nth term as a_n = a_1 · r^n instead of a_1 · r^(n−1), and in a sum uses n terms when the problem counts from the second term onward.
The formula is recalled as a shape rather than derived, so the exponent is off by one. Word problems that start 'after the first year' or 'from the second term' shift the index, and the candidate substitutes the visible number without adjusting for the offset.
What to do instead
Always write the first three terms explicitly from the given data before applying any formula, then check that a_3 from the formula equals the third term written by hand. For sums, count the terms on the fingers of the written list. Derive a_n = a_1 r^(n−1) from the list once per practice session so the exponent is understood, not remembered.
When finding the intersection of a line and a circle, the candidate substitutes, obtains a quadratic, and reports only the discriminant or only one root; or concludes 'no intersection' from a negative discriminant without checking the arithmetic of the substitution.
The algebra is treated as the goal rather than as a tool for a geometric answer, and the candidate stops as soon as a number appears. Sign errors in expanding (x − 2)² are common and are not caught because the result is never interpreted geometrically.
What to do instead
Set out a fixed three-line protocol: substitute to get the quadratic, compute the discriminant and state what it means geometrically (two points, one tangent point, no intersection), then solve fully for both coordinates including y. Finish by checking that each point satisfies both the line and the circle equation.
In probability, the candidate adds probabilities for events that are not mutually exclusive, or multiplies probabilities for events that are not independent, e.g. computing P(draw two aces without replacement) as (4/52)·(4/52).
The addition and multiplication rules are learned as slogans without their conditions. Drawing without replacement changes the sample space, but the candidate does not re-read the sampling description once the numbers are in hand.
What to do instead
Before computing, write one sentence describing the experiment: 'drawing without replacement, so the second draw depends on the first'. Then choose the rule from that sentence, not from the numbers. Use a tree diagram with the second-level denominators reduced by one, and check that all branch probabilities at each node sum to 1.
In calculus, the candidate differentiates a product as f′·g′ or a quotient as f′/g′, and finds maxima or minima by setting the derivative to zero without checking the sign change or the endpoints of a closed interval.
The power rule is over-generalised to every operation because it is the first and most practised rule. The second-derivative or sign-table check is skipped because the equation f′(x) = 0 already produces a number that looks like an answer.
What to do instead
Write the rule name beside each derivative ('product rule', 'quotient rule') and state the rule's formula before applying it. For optimisation, always build a sign table for f′ across the critical points and evaluate f at the interval endpoints, then state the conclusion in words ('maximum value 12 at x = 3 on [0, 5]').